Answer
The series converges to $5$, that is
$\mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( {\ln 5} \right)}^n}}}{{n!}} = 5$
Work Step by Step
From Table 2 we have
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting $x = \ln 5$, we get
${{\rm{e}}^{\ln 5}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( {\ln 5} \right)}^n}}}{{n!}}$
Since ${{\rm{e}}^{\ln 5}} = 5$, so the series converges to $5$, that is
$\mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( {\ln 5} \right)}^n}}}{{n!}} = 5$
